Linked Questions

128
votes
6answers
29k views

Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?

It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
40
votes
4answers
4k views

Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

I have read the following theorem: If $p_1,p_2,\dots,p_n$ are distinct prime numbers, then$$\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.$$ I have tried to prove a ...
32
votes
3answers
4k views

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[...
27
votes
3answers
2k views

Is it possible for integer square roots to add up to another?

I initially was wondering if it were possible for there to be three $x,y,z \in \mathbb{Q}$ and $\sqrt{x},\sqrt{y},\sqrt{z} \notin \mathbb{Q}$ such that $\sqrt{x} + \sqrt{y} = \sqrt{z}$. I had ...
26
votes
5answers
5k views

Can a finite sum of square roots be an integer? [duplicate]

Can a sum of a finite number of square roots of integers be an integer? If yes can a sum of two square roots of integers be an integer? The square roots need to be irrational.
21
votes
6answers
4k views

Is there a way to write an infinite set that contains only irrational numbers without integer multiples?

Is there a way to write an infinite set that contains only irrational numbers without integer multiples? The infinite set must not contain integer multiples of any other members of that set. For ...
19
votes
3answers
5k views

How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$?

How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field. I want to know the detail about the proof. Thanks in advance....
13
votes
5answers
9k views

Infinite Degree Algebraic Field Extensions

In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; | \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example ...
13
votes
2answers
259 views

Proving that if $x_1,\dots,x_n$ are rational numbers and $\sqrt{x_1}+\dots\sqrt{x_n}$ is rational, then each $\sqrt{x_i}$ is rational as well

I'm having a hard time with the following problem: Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$ is rational, then all $\sqrt{x_i}$ are ...
10
votes
5answers
220 views

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Question: Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. Following from the question, I tried: ...
9
votes
2answers
781 views

Show that $ a,b,c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $

Assume that $a,b,c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational,prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational. I know that can be proved, would like to know that there is ...
9
votes
2answers
96 views

Are square root binomials unique?

In Euclid we find the notion of a binomial, its simply a sum $s = \sqrt{a}+\sqrt{b}$ of two square roots $\sqrt{a}$ and $\sqrt{b}$. Lets say such a sum is simple iff $a$ and $b$ are positive non-zero ...
7
votes
2answers
1k views

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$?

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | ...
7
votes
2answers
132 views

Is $(5+\sqrt[3]2)^n$ ever an integer for $n \in \Bbb Z \setminus \{0\}$?

In general, I would like to prove that if $m>2$ is an integer, then $(5+\sqrt[m]2)^n$ is never an integer (unless for $n=0$). First, I'm interested in the simple case $m=3$ (I already solved it ...
7
votes
1answer
544 views

Is the sum of the square roots of all natural numbers up to n whole for any value of n other than 1?

For the summation $\sum_{n=0}^x \sqrt{n}$ are there any values of $x$ where the summation equals a whole number other than 1?
6
votes
1answer
5k views

Showing field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb{Q}$ degree 8 [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I am trying to classify the Galois group of the field extension $\mathbb{Q}(\sqrt{2}, \sqrt{...
6
votes
1answer
501 views

Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers. [duplicate]

Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite ...
6
votes
1answer
97 views

Sum of 5 square roots equals another square root. What is the minimum possible value of the summed square root?

In the equation $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e}=\sqrt{f}$, each variable is a distinct positive integer. What is the least possible value for $f?$ Out of purely trial and error, I have ...
6
votes
1answer
614 views

Elementary proof that finite sums of square roots of primes is irrational

It is relatively easy to show that if $p_1$, $p_2$ and $p_3$ are distinct primes then $\sqrt{p_1}+\sqrt{p_2}$ and $\sqrt{p_1}+\sqrt{p_2}+\sqrt{p_3}$ are irrational, but the only proof I can find that $...
5
votes
7answers
665 views

Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals

I was trying to prove that $\sqrt[3] {2} ,\sqrt[3] {4}$ and $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was ...
5
votes
6answers
655 views

Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?

This isn't a homework question but one I found online. Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? We just started doing field theory in my class and I want extra practice, but I ...
4
votes
3answers
857 views

The sum of square roots of non-perfect squares is never integer [duplicate]

My question looks quite obvious, but I'm looking for a strict proof for this: Why can't the sum of two square roots of non-perfect squares be an integer? For example: $\sqrt8+\sqrt{15}$ isn't an ...
4
votes
2answers
1k views

Linear independence of roots over Q

Let $p_1,\ldots,p_k$ be $k$ distinct primes (in $\mathbb{N}$) and $n>1$. Is it true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in $\mathbb{R}^+$) ...
4
votes
5answers
114 views

Let $a_0+a_1x+…+a_nx^n$ be a non zero polynomial with integer coefficients.if $p(√2+√3+√6)=0$, the smallest possible value of n is?

Question Let $a_0+a_1x+....+a_nx^n$ be a non zero polynomial with integer coefficients.if $p(√2+√3+√6)=0$, the smallest possible value of n is? Honestly I have no idea how to begin to solve this ...
4
votes
4answers
84 views

Are these two fields the same?

I wanted to know if the field $\mathbb{Q}(i\sqrt{7}) = \mathbb{Q}(\sqrt{7}, i)$ are the same. I don't think they are because $i \notin \mathbb{Q}(i\sqrt{7})$?
4
votes
1answer
247 views

Infinitely many transcendental numbers over Q

My previous question was not well-framed so I will ask again: Can you explicitly produce an infinite set of real numbers which is algebraically independent over $\mathbb Q$?
4
votes
1answer
2k views

Finding a primitive element for the field extension $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$

Let $p_1,\ldots,p_n\in\mathbb{N}$ be different prime numbers, it can be shown that $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]=2^n$ and in any case it is clearly finite since $[\mathbb{Q}(\...
4
votes
1answer
365 views

Is there a way to show the sum of any different square root of prime numbers is irrational? [duplicate]

Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational ...
4
votes
1answer
188 views

$\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$ is Irrational

If $m_1 , m_2, \cdots m_n$ are natural numbers where at least one of them is not a perfect square, then how do I prove that the sum $$\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$$ is irrational? I'm ...
3
votes
3answers
4k views

Finding a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.

I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$. I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$. Hence, the basis should be $1,(\...
3
votes
3answers
279 views

Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? [duplicate]

Here $n_i$ are integral numbers, and not all of them are zero. It is natural to conjecture that similar statement holds for even more prime numbers. Namely, $$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \...
3
votes
1answer
130 views

Show that $\sqrt{3}$ is not an element of the $\text{span} (1, \sqrt{2})$

So we have the following given to us this weekend just on a handout: Considering $\mathbb R$ as a vector space over the field $\mathbb Q$, show that $\sqrt 3$ is not an element of the span of $(1, \...
3
votes
5answers
167 views

show that $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$

I'd like to prove that $\sqrt{3}$ is not in the field $\mathbb{Q}(\sqrt{2})$. Let's write out a system of equations: $$ x^2 = 3 y^2 \text{ with } x,y \in \mathbb{Z}[\sqrt{2}] $$ Then if we write $x =...
3
votes
3answers
127 views

$[E:\mathbb{Q}]$, the degree of $E$ over $\mathbb{Q}$

Algebraic Extension $E=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, I need to find the $[E:\mathbb{Q}]$, the degree of $E$ over $\mathbb{Q}$ I know degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $...
3
votes
2answers
279 views

Algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Alternative proof?

Let $\mathcal{A}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Prove that $[\mathcal{A}:\mathbb{Q}] = \infty$. I can show this using $[\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}] = n$ for all $n ...
3
votes
2answers
151 views

How do I show that $\sqrt{5}\notin \mathbb{Q}(\sqrt{2},\sqrt{3})$?

How do I show that $\sqrt{5}\notin \mathbb{Q}(\sqrt{2},\sqrt{3})$? Since $X^2-5$ is the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ and its degree is not relatively prime to $[\mathbb{Q}(\sqrt{...
3
votes
3answers
88 views

Irrationality of $(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$

Sorry, for a rather silly question. Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square. ...
3
votes
1answer
226 views

Sum of square root of non perfect square positive integers is always irrational?

Let $S$ be a set of positive integers such that no element of $S$ is a perfect square. Is it true that $\sum_{s_i \in S} \sqrt{s_i}$ is always irrational? Motivation. Suppose the length of the ...
3
votes
0answers
73 views

Linear independence in $\mathbb{R}$ [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I would like to prove that the family $\{\sqrt{p}, p\text{ prime number} \}$ is linearly ...
2
votes
3answers
455 views

Showing that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$

How can I prove that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ ? I can only think of trying to write $\sqrt5 = a+b\sqrt7$ (where $a,b$ are in $\mathbb{Q}$), but I can't think of a good reason that ...
2
votes
2answers
115 views

Prove that $[ \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=8.$

I have to solve the following exercise: Compute $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]$ and $\operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}).$ Here my attempt: ...
2
votes
2answers
57 views

What are the applications of the result that $x$ is not a square in $F(x)$?

Let $F(x)$ denote the field of quotients of the ring $F[x]$. Prove that there is no element in $F(x)$ whose square is $x$. Solution:If possible let $\frac{f(x)}{g(x)} \in F(x),g(x)\neq 0$ such ...
2
votes
1answer
188 views

Prove that $[\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n}):\mathbb{Q}]=r^n$

We have $n$ distinct prime numbers $p_1,\cdots ,p_n$ and I am asked to show that $[\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n}):\mathbb{Q}]=r^n$ where $r\in \mathbb{N}$. I tried to solve it by ...
2
votes
1answer
212 views

Dimension of the algebraic closure of a continuum field of characteristic zero

Let me start by saying that I have no idea in algebra/number theory/whatever, so, please, forgive my ignorance. Let $\mathbb{F}$ be a field of characteristic zero and continuum cardinality, which ...
2
votes
1answer
505 views

sum of square root of primes 2

I dont know how to solve the problem below. (1) $p[1]$, $p[2]$, $\ldots$, $p[n]$ are distinct primes, where $n = 1,2,\ldots$ Let $a[n]$ be the sum of square root of those primes, that is, $a[n] = \...
2
votes
1answer
94 views

For $n\ge 3, x_{1},…,x_{n} \in \mathbf{Q}^{\ast}$, $[\mathbf{Q}(\sqrt{x_{1}},…\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$

For $n\ge 3, x_{1},...,x_{n} \in \mathbf{Q}^{\ast}$ and $[\mathbf{Q}(\sqrt{x_{1}},...\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$ how can we conclude that there are non empty $I \subset \{1,...,n\}$ with $...
2
votes
2answers
97 views

How to prove the set $\{\sqrt{n}:\textrm{$n$ is squarefree}\}$ to be a linearly independent set?

As the title goes, I am stuck on this problem. Prove that the set $\{\sqrt{n}:\textrm{$n$ is squarefree}\} =\{1,\sqrt{2}, \sqrt{3},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{10},\ldots\}$ is a linearly ...
2
votes
1answer
310 views

$\sqrt{p_1}$ is not in $Q[\sqrt{p_2},…,\sqrt{p_n}]$ [duplicate]

How to show $\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$ if $p_1,...,p_n$ are distinct primes? Intuitively, this is pretty clear, but it makes me very uncomfortable to just believe. Any idea ...
2
votes
1answer
69 views

Splitting field of $(x^2-2)(x^6-20)$ over $\mathbb{Q}$

I have to determine the splitting field $K$ of $f(x)=(x^2-2)(x^6-20)$ over $\mathbb{Q}$. My attempt of solution: $K=\mathbb{Q}(\sqrt2, \sqrt[6]{20}, i\sqrt3)$; $d_1:=[\mathbb{Q}(\sqrt2, \sqrt[6]{20})(...
2
votes
1answer
146 views

Exercise about field extensions [duplicate]

Consider $a_1,\ldots,a_n\in \mathbb Z$. i) Suppose $a_1,\ldots, a_n$ are pairwise relatively prime. I have to see by induction on n that $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q]=2^n$ Once ...

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